Problem 2: Let X₁,…, Xn be i.i.d. random variables from a Geometric(p), p = (0, 1), which means that Px(k; p) = (1 − p)k−¹p for k = 1, 2, and we have E[X;] = 1/p. (a) Find the asymptotic variance of the maximum likelihood estimator î. (b) Let n = 100 and you observe the following data: Number 1 2 3 4 5 Frequency 60 21 12 4 3 which means that 60 of the random variables X₁,…, X100 equal 1, 21 of them equal 2, and so on. Find a 95% confidence interval for p.
Problem 2: Let X₁,…, Xn be i.i.d. random variables from a Geometric(p), p = (0, 1), which means that Px(k; p) = (1 − p)k−¹p for k = 1, 2, and we have E[X;] = 1/p. (a) Find the asymptotic variance of the maximum likelihood estimator î. (b) Let n = 100 and you observe the following data: Number 1 2 3 4 5 Frequency 60 21 12 4 3 which means that 60 of the random variables X₁,…, X100 equal 1, 21 of them equal 2, and so on. Find a 95% confidence interval for p.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![Problem 2: Let X₁,..., Xn be i.i.d. random variables from a Geometric (p), p = (0,1),
which means that Px(k; p) = (1 − p)k-¹p for k = 1, 2, ... and we have E[X;] = 1/p.
(a) Find the asymptotic variance of the maximum likelihood estimator p.
(b) Let n = 100 and you observe the following data:
Number
12 3 4 5
Frequency 60 21 12 4 3
which means that 60 of the random variables X₁,..., X100 equal 1, 21 of them equal 2,
and so on. Find a 95% confidence interval for p.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F393ac1f3-b11c-4a5d-aefe-7fd8292ae47c%2F411ee2d4-d38f-43c5-9c9f-92f8ce7da740%2Fzr17rw_processed.png&w=3840&q=75)
Transcribed Image Text:Problem 2: Let X₁,..., Xn be i.i.d. random variables from a Geometric (p), p = (0,1),
which means that Px(k; p) = (1 − p)k-¹p for k = 1, 2, ... and we have E[X;] = 1/p.
(a) Find the asymptotic variance of the maximum likelihood estimator p.
(b) Let n = 100 and you observe the following data:
Number
12 3 4 5
Frequency 60 21 12 4 3
which means that 60 of the random variables X₁,..., X100 equal 1, 21 of them equal 2,
and so on. Find a 95% confidence interval for p.
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